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We are given the equation:

\[
\frac{(n+5)!}{n!(n+4)} = \frac{9!}{3!}
\]

Our goal is to find the integer \( n \) that satisfies this equation.

### Step 1: Simplify the equation
Start by simplifying both sides of the equation:

The right-hand side is:
\[
\frac{9!}{3!}
\]
Calculate the factorials:

\[
9! = 362880, \quad 3! = 6
\]
Thus,
\[
\frac{9!}{3!} = \frac{362880}{6} = 60480
\]

Now the equation becomes:
\[
\frac{(n+5)!}{n!(n+4)} = 60480
\]

### Step 2: Simplify the left-hand side
We can rewrite \( (n+5)! \) as:
\[
(n+5)! = (n+5)(n+4)(n+3)(n+2)(n+1)n!
\]
Substitute this into the left-hand side:
\[
\frac{(n+5)(n+4)(n+3)(n+2)(n+1)n!}{n!(n+4)} = 60480
\]
Cancel out \( n! \) and \( (n+4) \) from the numerator and denominator:
\[
(n+5)(n+3)(n+2)(n+1) = 60480
\]

### Step 3: Solve for \( n \)
Now we solve the equation:
\[
(n+5)(n+3)(n+2)(n+1) = 60480
\]

We can try substituting different values of \( n \) to find which satisfies this equation.

Let me calculate that.From the calculations, we can see that when \( n = 13 \), the left-hand side equals 60480, which matches the right-hand side of the equation.

Thus, the integer \( n \) that satisfies the equation is:
\[
n = 13
\]

Would you like more details or have any questions? Here are 5 related questions to explore:

1. How do factorials grow as \( n \) increases?
2. What are the properties of factorials in algebraic expressions?
3. Can we solve other factorial equations using a similar method?
4. How does trial and error work in finding integer solutions to equations?
5. What are some practical applications of factorials?

**Tip:** When solving factorial equations, it's helpful to simplify and cancel common terms early to make the calculations more manageable.

Find an integer n satisfying \( \frac{(n+5)!}{n!(n+4)} = \frac{9!}{3!} \)

Learn how to solve the factorial equation \( \frac{(n+5)!}{n!(n+4)} = \frac{9!}{3!} \) step by step to find the integer n. This problem involves simplifying factorials and solving algebraic equations, suitable for high school students in Grades 9-12.

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